UVA - 10404 Bachet's Game
2013-09-13 21:01
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题意:给n个石头,m 种取法,Stan先取,Ollie后取,取到最后一块的获胜,问谁赢了,首先我们想想怎么能让Stan获胜,一定是Ollie取完后,我们可以通过m种取法中的一种使得Stan拿到最后一块,那么我们用dp[i] = 1表示Stan赢,0为输,那么一定有dp[n-arr[i] ] == 0,那么Stan 就赢了,推广到一般的情况是只要dp[i-arr[j]] == 0,那么我们就可以说在i个石头的情况下Stan赢了,这样我们就得到了状态转移方程
if(i>=arr[j] && dp[i-arr[j]]==0)dp[i]=1; 所以我们第一块石头开始判断,只要dp
== 1 ,那么Stan 就是赢家
if(i>=arr[j] && dp[i-arr[j]]==0)dp[i]=1; 所以我们第一块石头开始判断,只要dp
== 1 ,那么Stan 就是赢家
#include <iostream> #include <cstdio> #include <cstring> using namespace std; const int MAXN = 1000005; int dp[MAXN]; int arr[MAXN]; int n,m; int main(){ while (scanf("%d%d",&n,&m) != EOF){ memset(dp,0,sizeof(dp)); for (int i = 1; i <= m; i++){ scanf("%d",&arr[i]); dp[arr[i]] = 1; } for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) if (i >= arr[j] && !dp[i-arr[j]]){ dp[i] = 1; break; } if (dp ) printf("Stan wins\n"); else printf("Ollie wins\n"); } return 0; }
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